3.8 Wilcoxon- Mann- Whitney Test

For two independent samples, the analogue of the one sample. A statistic \(U\), a function of the rank sum \(S\), can be calculated for either groups in order to determine the strength of the evidence against \(H_0\) for the first of two samples this statistic is given by \[U_n=S_n-\frac {n(n+1)}{2}\] 2 equivalent statistic from the other sample \(\displaystyle {U_m=S_m-\frac {m(m+1)}{2}}\)
We only need to compute one of \(S_n/S_m\), for sum all ranks from 1 to \(N_0=(m+n)\) is \[S=\frac {1}{2}N(N+1)\hspace {0.5cm} \text {where}\hspace {0.4cm} N=n+m\] it can be shown that \(\hspace {0.2cm} U_m =mn-U_n\hspace {0.4cm} \text {or}\hspace {0.4cm} U_n =mn-U_m\)

For large samples \((n=m=16)\) test statistic is \begin {align*} E(U/H_0) &=\frac {mn}{2}\\\\ \text {var}(U/H_0) &=\frac {mn(m+n+1)}{12}\\\\ \therefore \hspace {0.4cm} Z &=\frac {U-\dfrac {mn}{2}}{\sqrt {\dfrac {mn(m+n+1)}{12}}}\thicksim N(0,1) \end {align*}

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